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276Testing_and Confidence Regions Chizar [(jii) Consider the model 0Falv)0 # 0 "1 Foly), 0 = 0.To see whether the new treatment is beneficial, we test H : 0 &l...

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276Testing_and Confidence Regions Chizar [(jii) Consider the model 0Falv)0 # 0 "1 Foly), 0 = 0.To see whether the new treatment is beneficial, we test H : 0 < versus k 0 > U. Assume that Fo has density foly) . Show that the UMP test is based on the slatistic 5RLI Fo(Y) 9/Let Xi beiid with distribution function Flr) . We want test whether F is exponential, F(c) = 1 exp(~r)r > 0, or Weibull, F(r) = 0 > 0. Find the MP test for testing H : 0 exp(-I" ) not UMP versus K 01 > S

276 Testing_and Confidence Regions Chizar [ (jii) Consider the model 0Falv) 0 # 0 "1 Foly), 0 = 0. To see whether the new treatment is beneficial, we test H : 0 < versus k 0 > U. Assume that Fo has density foly) . Show that the UMP test is based on the slatistic 5RLI Fo(Y) 9/Let Xi beiid with distribution function Flr) . We want test whether F is exponential, F(c) = 1 exp(~r)r > 0, or Weibull, F(r) = 0 > 0. Find the MP test for testing H : 0 exp(-I" ) not UMP versus K 01 > Show that the lestis Show that under the assumptions of ncotet complete _ 4.3.2 the class of all Bayes tests Hint: Consider the class of all Bayes tests of H "{00 } x{01 } varies between Qw versus K when' and Show that under the assumptions of Teonete H : Oo versus K 01 is of the fori L3Land 0-1 loss, every Bayes test for for some Kint: Bayes lest rejects (accepts) H if p(z, O)da(0) / P(z, O)d(0) (<) 1. The left-hand side equals L(T; Oo )dn(0) L(e, Vo )dg(0) The numerator an increasing funclion of T(z). the denominator decreasing- 12 . Show that under the assumptions of Theorem 4..1; IUMP 0o versus K esliqg H : 0 2 Froblems for Section 4.4 Let sample from oma population with uunknor Glu;t



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Consider testing hypotheses $H_{0} : p=p_{0}$ versus $H_{a} : p<p_{0} .$ Suppose that, in fact, the true value of the parameter $p$ is $p^{\prime},$ where $p^{\prime}<p_{0}\left($ so $H_{\text { a }}$ is true). \right.(a) Show that the expected value and variance of the test statistic $Z$ in the one-proportion
$z$ test are
$E(Z)=\frac{p^{\prime}-p_{0}}{\sqrt{p_{0}\left(1-p_{0}\right) / n}}$
$\operatorname{Var}(Z)=\frac{p^{\prime}\left(1-p^{\prime}\right) / n}{p_{0}\left(1-p_{0}\right) / n}$
(b) It can be shown that $P$ -value $\leq \alpha$ iff $Z \leq-z_{\alpha},$ where $-z_{\alpha}$ denotes the $\alpha$ quantile of the standard normal distribution (i.e., $\Phi\left(-z_{\alpha}\right)=\alpha ) .$ Show that the power of the lower-tailed one-sample $z$ test when $p=p^{\prime}$ is given by
$\Phi\left(\frac{p_{0}-p^{\prime}-z_{\alpha} \sqrt{p_{0}\left(1-p_{0}\right) / n} {\sqrt{p^{\prime}\left(1-p^{\prime}\right) / n}}\right)$
(c) A package-delivery service advertises that at least 90$\%$ of all packages brought to its office
by 9 a.m. for delivery in the same city are delivered by noon that day. Let $p$ denote the true
proportion of such packages that are delivered as advertised and consider the null
hypotheses $H_{0} : p=.9$ versus the alternative $H_{\mathrm{a}} : p<9 .$ If only 80$\%$ of the packages are delivered as advertised, how likely is it that a level. 01 test based on $n=225$ packages will detect such a departure from $H_{0} ?$

We want to conduct a p. D. T. A pair of differences to us at the alpha equals 1% level passing the claim that X does not equal XB. Or that our population means are not equal. We have the data A be given here and we assume the data has a mound shaped symmetric distribution on the right. Have already concluded or rather computed the differences mean. D. Bar sample size, N. S. D. 3.75 So we proceed to the following five steps below. To complete this problem first we evaluate the requirements hypotheses because of the distribution shape. It is appropriate to determine the distribution. We've degree of freedom and minus 2016. No hypothesis mute equals zero. Alternative hypothesis clearly does not equal zero and confidence 00.1 So artist, artistic and P value as follows. Two is equal to D. Bar over SDI or 10 or 1.17 from the tea table. This gives p between 0.5 point 25 Thus we conclude P is greater than alpha. We fail to reject the hypothesis, meaning we lack evidence that nearly does not equal zero

We want to conduct a pair differences test at the alpha equals 5% level testing the claim that population mean X bar A is greater than population X barbie. We have data a be given here, we assume amounts to mr distribution as you can see on the right. I've already calculated the mean difference D bar 6.125 The sample size and eight and the sample standard deviation of differences SD and 8.7 We complete the five steps us to blow to solve this problem first, let's evaluate the requirements to use a student's T distribution of the hypotheses because of the distribution shape it is appropriate to use a student distribution your degree of freedom and minus 27. No hypothesis mute equals zero. Alternative media is greater than zero and alpha equals 00.5 for confidence nexus, complete the test at and the P value our test that is T equals D. Bar over SDR. Again this gives 2.14 U. T. Table. This gives us a P value between 0.5 point 025 That means we can include that P is less than equal to alpha. So we reject the null hypothesis H not which means that we have evidence and you D. Is greater than zero.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.


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